is now 12.5. This means that xi — Xi — X is unaffected. Hence ‘Jf xi2 is the

i=i

same and since Yi, is unchanged, we conclude that "ols is still the same at

0.8095. However, aols = Y — "olsX is changed because X is changed. This is now aols = 6.5 — (0.8095/(12.5) = —3.6188. It has decreased by 5"ols since X increased by 5 while "ols and Y remained unchanged. It is easy to see that Yi = aols + "olsXi remains the same. When Xi increased by 5, with "ols the same, this increases Yi by 5"ols. But aols decreases Yi by —5"ols. The net effect on Yi is zero. Since Yi is unchanged, this means ei = Yi — Yi

b. Pr[0.1 < X < 0.3] = /У dx = 0.3 — 0.1 = 0.2. It does not matter if we include the equality signs Pr[0.1 < X < 0.3] since this is a continuous random variable. Note that this integral is the area of the rectangle, for X between 0.1 and 0.3 and height equal to 1. This is just the length of this rectangle, i. e., 0.3 — 0.1 = 0.2.

The income elasticity is —0.227 which is negative! Its standard error is (0.1999) and the t-statistic fortesting this income elasticity is zero is —1.135 which is insignificant with a p-value of 0.26. Hence, we cannot reject the null hypothesis. R2 = 0.0285 and s = 0.19085. This regression is not very useful...